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What will be the temperature at which a solution containing 6 g of glucose per 1000 g water with boil, if molal elevation constant for water is 0.52/1000 g. |
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Answer» `1000.173^(@)`C `DeltaT_(b)=(1000xxK_(b)xxw)/(180xx1000)=0.0173^(@)C` HENCE boiling point of solution =b.p of WATER + `DeltaT_(b)=100+0.0173^(@)C` |
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